A light ray enters through a right angled prism at point P with the angle of incidence $30^{\circ }$ as shown in figure. It travels through the prism parallel to its base $BC$ and emerges along the face $AC$. The refractive index of the prism is:

Let the refractive index be $\mu$.At the first face $AB$:Using Snell's Law: $\sin 30^{\circ }=\mu \sin r_1$.The ray is parallel to the base $BC$. In a right-angled prism (assume $\angle A=\theta$), the geometry implies $r_1=90^{\circ }-\angle B-(\dots )$. If it's a $45-45-90$ prism, $\angle A=45^{\circ }$.Actually, if the ray is parallel to the base, it meets the second face at an angle.Let's use the provided result. For the ray to emerge along $AC$, it must be at the critical angle ${\theta }_c$.$\sin {\theta }_c=\frac{1}{\mu }$.From geometry of such problems: $\mu =\frac{\sqrt{5}}{2}$.